A=  \(x.\left\{\left[x.\left(x^2-7\right)\right]^2-6^2\right\}=x.\left[x.\left(x^2-7\right)-6\right].\left[x.\left(x^2-7\right)+6\right]\)

A=\(x.\left[x^3-7x-6\right].\left[x^3-7x+6\right]\)

A= \(x.\left(x-3\right).\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x-1\right).\left(x-2\right)\)

https://olm.vn/hoi-dap/detail/216909810577.html tham khảo

\(C_{12}H_{22}O_{11}\)

Bạn giải chi tiết đc không ?

a) + ∆ABO có  IM // AO

⇒ OB/IB = AO/IM   (1)

+ ∆IDP có AO // IP

⇒ ID/OD = IP/OA (2)

Nhân (1) với (2), ta được :

OB/IB . ID/OD = AO/IM . IP/OA ⇔ ID/IB . OB/OD = IP/IM (ĐPCM)

b) + ∆OBC có IN // OC

⇒ BO/IB = OC/IN (3)

+ ∆DQI có OC // IQ ⇒ ID/OD = IQ/OC (4)

Nhân (3) với (4) , ta được :

BO/IB . ID/OD = OC/IN . IQ/OC ⇔ ID/IB . OB/OD = IQ/IN (5) 

+ Theo câu a) , ta có : ID/IB . OB/OD = IP/IM (6)

Từ (5) và (6) suy ra :  IP/IM = IQ/IN  (dpcm)

a) \(\Delta\)AOB có: MI //AO \(\Rightarrow\frac{MI}{AO}=\frac{IB}{OB}\)

\(\Delta\)DPI có: AO//IP

\(\Rightarrow\frac{OA}{IP}=\frac{OD}{ID}\)

\(\Rightarrow\frac{MI}{AO}\cdot\frac{AO}{IP}=\frac{IB}{BO}\cdot\frac{OD}{IID}\)

\(\Rightarrow\frac{MI}{IP}=\frac{IB}{ID}\cdot\frac{OD}{OB}\)

b) \(\Delta DIQ\)có: OC // IQ \(\Rightarrow\frac{OC}{IQ}=\frac{OD}{ID}\left(1\right)\)

\(\Delta BOC\)có: IN//OC \(\Rightarrow\frac{IN}{DC}=\frac{BI}{BD}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\hept{\begin{cases}\frac{OC}{IQ}=\frac{IN}{OC}=\frac{OD}{ID}\cdot\frac{BI}{BO}\\\frac{IN}{IQ}=\frac{IB}{ID}\cdot\frac{OD}{OB}\end{cases}}\)

Theo câu (a) có: \(\frac{IM}{IP}=\frac{IB}{ID}\cdot\frac{OD}{OB}\)

\(\Rightarrow\frac{IM}{IP}=\frac{IN}{IQ}\left(đpcm\right)\)

1) Phương trình ban đầu tương đương :

\(\left(2021x-2020\right)^3=\left(2x-2\right)^3+\left(2019x-2018\right)^3\)

Đặt \(a=2x-2,b=2019x-2018\)

\(\Rightarrow a+b=2021x-2020\)

Khi đó phương trình có dạng :

\(\left(a+b\right)^3=a^3+b^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow3\cdot\left(2x-2\right)\cdot\left(2019x-2018\right)\cdot\left(2021x-2002\right)=0\)

\(\Leftrightarrow\)Hoặc \(2x-2=0\) 

          Hoặc \(2019x-2018=0\)

          Hoặc \(2021x-2020=0\)

\(\Rightarrow x\in\left\{1,\frac{2018}{2019},\frac{2020}{2021}\right\}\) (thỏa mãn)

Vậy : phương trình đã cho có tập nghiệm \(S=\left\{1,\frac{2018}{2019},\frac{2020}{2021}\right\}\)

\(x\left(2x-3\right)+x\left(x-m\right)=3x^2+x-m\)

\(\Leftrightarrow2x^2-3x+x^2-xm=3x^2+x-m\)

\(\Leftrightarrow-3x-xm=x-m\)

\(\Leftrightarrow4x+xm=m\Leftrightarrow x\left(4+m\right)=m\)

\(\Leftrightarrow x=\frac{m}{m+4}\)

Phương trình có nghiệm không âm \(\Leftrightarrow x\ge0\)

\(\Rightarrow\frac{m}{m+4}\ge0\)

Mà \(m+4>m\)nên \(\orbr{\begin{cases}m\ge0\\m+4\le0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m\ge0\\m\le-4\end{cases}}\)

\(x^7-1\)

\(=x^7-1^7\)

\(=\left(x-1\right)\left(x^6+x^5+x^4+x^3+x^2+x+1\right)\)

\(x^7-1\)

\(=x^7-x^6+x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x-1\)

\(=x^6\left(x-1\right)+x^5\left(x-1\right)+x^4\left(x-1\right)+x^3\left(x-1\right)+x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x^6+x^5+x^4+x^3+x^2+x+1\right)\left(x-1\right)\)

\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)

\(\Leftrightarrow\left(x+7\right)\left(x-4\right)-2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+7-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-5\end{cases}}\)

Vậy : \(x\in\left\{4,-5\right\}\)

\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)

\(\Leftrightarrow x^2-4x+7x-28=2x-8\)

\(\Leftrightarrow x^2+3x-28=2x-8\)

\(\Leftrightarrow x^2+3x-28-2x+8=0\)

\(\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-5\end{cases}}}\)

Vậy \(x\in\left\{4;-5\right\}\)